# For which values of $a\in\mathbb ℤ/3\mathbb ℤ$ is the quotient $\mathbb ℤ/3\mathbb ℤ[x]/(x^3+x^2+ax+1)$ a field?

I'm trying to solve the following problem:

Determine for which values of $a\in\mathbb{Z}/3\mathbb{Z}$ the quotient $Q_a=(\mathbb{Z}/3\mathbb{Z})[x]/(x^3+x^2+ax+1)$ is a field.

I see two options:

• Show that ($x^3+x^2+ax+1$) is maximal, or
• show that every element of $Q_a \backslash \{0\}$ is invertible.

Any help would be appreciated.

• You can try finding for which values of $a$ the polynomial has a root. This will tell whether or not it is irreducible, as it has degree 3. Aug 5, 2014 at 15:45
• Since $\mathbb Z/3\mathbb Z$ is a field, and $F[x]$ is a PID for any field, you only need to show that $x^3+x^2+ax+1$ is irreducible (or prove it isn't.) Aug 5, 2014 at 15:46

The first option looks good, and recall that here "maximal" means that the generating element, that is the polynomial, is irreducible over $\mathbb{Z}/3\mathbb{Z}$.

Now, you are left with the task of deciding which of the polynomials are irreducible.

I cannot know which means you have for this, but if nothing else you can note that since the polynomials have degree $3$ they are reducible if and only if they have a root. (If this is not clear, try to prove it.)

Then, you can just check which polynomial has a root, for instance by plugging in the three possible values.

Here is my solution, thanks for your help.

$$Q_a$$ is a field if and only if $$I:=(x^3+x^2+ax+1)$$ is maximal.

$$I$$ is maximal if and only if $$I=(p(x))$$ for $$p(x)\in(\mathbb{Z}/3\mathbb{Z})[x]$$ an irreducible polynomial.

If a polynomial of degree 2 or 3 has no roots in $$\mathbb{Z}/3\mathbb{Z}$$, then it is irreducible in $$(\mathbb{Z}/3\mathbb{Z})[x]$$.

$$f(x):= x^3+x^2+ax+1,\quad f(0)=1,\quad f(1)=a,\quad f(2)=2a+1$$

f is irreducible only for $$a=2$$ so only $$Q_2$$ is a field.

Tell me if something is wrong.

• I cannot find a mistake here. Aug 5, 2014 at 17:17